c Pleiades Publishing, Ltd., 2018. ISSN 2070-0466, p-Adic Numbers, Ultrametric Analysis and Applications, 2018, Vol. 10, No. 1, pp. 12–31. 

RESEARCH ARTICLES

New Applications of the p-Adic Nevanlinna Theory∗1 Alain Escassut1** and Ta Thi Hoai An2, 3*** 1

Universite´ Clermont Auvergne, Laboratoire de Mathematiques ´ Blaise Pascal, BP 10448, CNRS, UMR 6620, LMBP, F-63171 AUBIERE, France 2 Institute of Mathematics, Vietnam Academy of Science and Technology, 18, Hoang Quoc Viet, Hanoi, Vietnam 3 Institute of Mathematics and Applied Sciences (TIMAS), Thang Long University, Hanoi, Vietnam Received December 1, 2017

Abstract—Let IK be an algebraically closed field of characteristic 0 complete for an ultrametric absolute value. Following results obtained in complex analysis, here we examine problems of uniqueness for meromorphic functions having finitely many poles, sharing points or a pair of sets (C.M. or I.M.) defined either in the whole field IK or in an open disk, or in the complement of an open disk. Following previous works in C, l we consider functions f n (x)f m (ax + b), g n (x)g m (ax + b) f is a n + m-th root of 1 with |a| = 1 and n = m, sharing a rational function and we show that g whenever n + m ≥ 5. Next, given a small function w, if n, m ∈ IN are such that |n − m|∞ ≥ 5, then f n (x)f m (ax + b) − w has infinitely many zeros. Finally, we examine branched values for meromorphic functions f n (x)f m (ax + b). DOI: 10.1134/S2070046618010028 Key words: p-adic meromorphic functions, Nevanlinna’s theory, values distribution, small functions, Picard values, branched values.

1. INTRODUCTION Let IK be a complete ultrametric algebraically closed field of characteristic 0 whose ultrametric absolute value is denoted by | . |. The Nevanlinna theory, well known for complex meromorphic functions [17], was examined over IK by Ha Huy Khoai [10] and A. Boutabaa [4]. Next, in [5] a similar theory was made for unbounded meromorphic functions in an “open" disk of IK, taking into account Lazard’s problem. In [11], M. O. Hanyak and A. A. Kondratyuk constructed a Nevanlinna theory for meromorphic functions in a set of the form C l \ {a1 , ..., am }, where the meromorphic functions can admit essential singularities at a1 , ..., am . Here we recall the Nevanlinna theory for meromorphic functions in the complement of an open disk. Next, we can apply this to obtain results on uniqueness and branched values as it was done in similar problems [1, 2, 4, 5]. We then show new results on meromorphic functions sharing two sets and properties of meromorphic functions of the form f n (x)f m (ax + b) with regards to branched values and Picard’s values. All results of Paragraphs 1 to 5 were already published with all proofs in [2, 5, 8, 9, 15]. Therefore we will only give the proofs of theorems presented in Paragraph 6. Notation. Given r > 0, a ∈ IK we denote by d(a, r) the disk {x ∈ IK | |x − a| ≤ r}, by d(a, r − ) the disk {x ∈ IK | |x − a| < r}, and by C(a, r) the circle {x ∈ IK | |x − a| = r}. Given r  > r  , we put Δ(0, r  , r  ) = d(0, r  ) \ d(0, r − ). ∗

The text was submitted by the authors in English. This paper was presented by the first author at the Sixth International Conference on p-Adic Mathematical Physics and its Applications which was held in Mexico City, October 23-27, 2017. ** E-mail: [email protected] *** E-mail: [email protected] 1

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Henceforth, we fix R > 0, we denote by S the disk d(0, R− ) and put D = IK \ S. Given a subset E of IK having infinitely many points, we denote by R(E) the IK-algebra of rational functions h ∈ IK(x) having no pole in E. We then denote by H(E) the IK-vector space of analytic elements on E, i.e., the completion of R(E) with respect to the topology of uniform convergence on E. By classical properties of analytic elements in [8, 13], given a circle C(a, r) and an element f of +∞  cn (x − a)n converging whenever |x − a| = r, then |f (x)| H(C(a, r)), i.e., a Laurent series f (x) = −∞

is equal to sup |cn |r n in all classes of C(a, r) except maybe in finitely many. When a = 0, we put n∈ ZZ

|f |(r) = sup |cn |r n . n∈ ZZ

We denote by A(IK) the IK-algebra of entire functions in IK, by A(d(a, R− ) the IK-algebra of ∞  cn (x − a)n converging in all d(a, R− ) and by A(D) the IK-algebra of Laurent series power series ∞ 

n=0

cn (x − a)n converging in D. Similarly, we denote by M(IK) the field of meromorphic functions in IK,

−∞

i.e. the field of fractions of A(IK), by M(d(a, R− )) the field of meromorphic functions in d(a, R− ), i.e. the field of fractions of A(d(a, R− )), and by M(D) the field of meromorphic functions in D, i.e. the field of fractions of A(D). Next, we denote by Ab (d(a, R− )), the set of f ∈ A(d(a, R− )) that are bounded in d(a, R− ) and we put Au (d(a, R− )) = A(d(a, R− )) \ Ab (d(a, R− )). We denote by Mb (d(a, R− )) the field of fractions of Ab (d(a, R− )) and put Mu (d(a, R− )) = M(d(a, R− )) \ Mb (d(a, R− )). We denote by Az (D) the set of f ∈ A(D) admitting finitely many zeros in D and we put A∗ (D) = A(D) \ Az (D) and similarly, we denote by Mz (D) the field of fraction of Az (D) and we put M∗ (D) = M(D) \ Mz (D). So, M∗ (D) is the set of meromorphic functions in D having at least infinitely many zeros or infinitely many poles in D. 2. NEVANLINNA THEORY IN THE CLASSICAL p-ADIC CONTEXT Definition and notation. Let f ∈ M(d(a, R− )) (resp. f ∈ M(D)) and let α ∈ d(a, R− ), resp. α ∈ D). If f admits a zero of order q at α we set ωα (f ) = q and if f has no zero and no pole at α, we set ωα (f ) = 0. h Let f = ∈ M(d(a, R− )), (resp. f ∈ M(D)). For each α ∈ IK (resp. α ∈ d(a, R− ), resp. α ∈ D) l h the number ωα (h) − ωα (l) does not depend on the functions h, l chosen to make f = . Thus, we can l generalize the notation by setting ωα (f ) = ωα (h) − ωα (l). If ωα (f ) is an integer q > 0, α is called a zero of f of order q. If ωα (f ) is an integer q < 0, α is called a pole of f of order −q. If ωα (f ) ≥ 0, f is said to be holomorphic at α. Throughout the next paragraphs, we denote by I the interval [t, +∞[, by J an interval of the form [t, R[ with t > 0. We denote by f a function that belongs either to M(IK) or to M(S). We have to introduce the counting function of zeros and poles of f , counting or not multiplicity. Here we will choose a presentation that avoids assuming that all functions we consider admit no zero and no pole at the origin. Definitions. We denote by Z(r, f ) the counting function of zeros of f in d(0, r) in the following way. Let (an ), 1 ≤ n ≤ σ(r) be the finite sequence of zeros of f such that 0 < |an | ≤ r, of respective order sn . σ(r)  sn (log r − log |an |) and so, Z(r, f ) is called the countWe set Z(r, f ) = max(ω0 (f ), 0) log r + n=1

ing function of zeros of f in d(0, r), counting multiplicity. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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In order to define the counting function of zeros of f without multiplicity, we put ω0 (f ) = 0 if ω0 (f ) ≤ 0 and ω0 (f ) = 1 if ω0 (f ) ≥ 1.  σ(r)

Now, set Z(r, f ) = ω0 (f ) log r +

(log r − log |an |) and so, Z(r, f ) is called the counting function

n=1

of zeros of f in d(0, r) ignoring multiplicity. In the same way, considering the finite sequence (bn ), 1 ≤ n ≤ τ (r) of poles of f such that 0 < |bn | ≤ r, with respective multiplicity order tn , we put N (r, f ) = max(−ω0 (f ), 0) log r +

τ (r) 

tn (log r − log |bn |) and then N (r, f ) is called the counting func-

n=1

tion of the poles of f , counting multiplicity. Next, we put ω0 (f ) = 0 if ω0 (f ) ≥ 0 and ω0 (f ) = 1 if ω0 (f ) ≤ −1 and we set N (r, f ) = ω0 (f ) log r + τ (r)  (log r − log |bn |) and then N (r, f ) is called the counting function of the poles of f , ignoring n=1

multiplicity. Now, we can define the Nevanlinna function T (r, f ) in I or J as T (r, f ) = max(Z(r, f ), N (r, f )) and the function T (r, f ) is called characteristic function of f or Nevanlinna function of f . Finally, if Y is a subset of IK we will denote by Z Y (r, f  ) the counting function of zeros of f  , excluding those which are zeros of f − a for any a ∈ Y . Remark. If we change the origin, the functions Z, N, T are not changed, up to an additive constant. Theorem 2.1. Let f ∈ M(IK) (resp. f ∈ M(d(0, R− ))) have no zero and no pole at 0. Then log(|f |(r)) = log(|f (0)|) + Z(r, f ) − N (r, f ). Theorem 2.2: (First Main Theorem). Let f, g ∈ M(IK) (resp. f, g ∈ M(S)). Then Z(r, f g) ≤ Z(r, f ) +Z(r, g), N (r, f g) ≤ N (r, f ) + N (r, g), and T (r, f + b) = T (r, f ) + O(1), T (r, f g) ≤ T (r, f ) + 1 f T (r, g), T (r, f + g) ≤ T (r, f ) + T (r, g) + O(1), T (r, cf ) = T (r, f ) ∀c ∈ IK∗ , T (r, ) = T (r, f )), T (r, ) f g ≤ T (r, f )) + T (r, g). Let P (X) ∈ IK[X]. Then T (r, P (f )) = deg(P )T (r, f ) + O(1) and T (r, f  P (f )) ≥ T (r, P (f )). Suppose now f, g ∈ A(IK) (resp. f, g ∈ A(S)). Then Z(r, f g) = Z(r, f ) + Z(r, g), T (r, f ) = Z(r, f )), T (r, f g) = T (r, f ) + T (r, g) + O(1) and T (r, f + g) ≤ max(T (r, f ), T (r, g)). Moreover, if lim T (r, f ) − T (r, g) = +∞ then T (r, f + g) = T (r, f ) when r is big enough. r→+∞

Theorem 2.3. Let f ∈ M(IK). Then f belongs to IK(x) if and only if T (r, f ) = O(log r). Corollary 2.3.a. Let f ∈ M∗ (IK). Then f is transcendental over IK(x). Theorem 2.4. Let f ∈ M(S). Then f belongs to Mb (S) if and only if T (r, f ) is bounded in [0, R[. Corollary 2.4.a. Let f ∈ Mu (S). Then f is transcendental over Mb (S). Theorem 2.5. Let f ∈ M(IK) (resp. f ∈ M(S)). Then for all k ∈ IN∗ , we have N (r, f (k) ) = N (r, f ) + kN (r, f ) and Z(r, f (k) ) ≤ Z(r, f ) + kN (r, f ) − k log(r) + O(1). Theorem 2.6. Let f ∈ M(IK) (resp. f ∈ M(S)) and let a1 , ..., aq ∈ IK be distinct. Then (q − 1)T (r, f ) q ≤ max1≤k≤q j=1,j=k Z(r, f − aj ) + O(1). Remark. The last Theorem does not hold in complex analysis. Indeed, let f be a meromorphic function ex . Then Z(r, f − a) + Z(r, f − b) = 0. in C l omitting two values a and b, such as f (x) = x e −1 p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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We can now state the Second Main Theorem [4, 5, 8]. Theorem 2.7: (Second Main Theorem). Let α1 , ..., αq ∈ IK, with q ≥ 2, let Y = {α1 , ..., αq } and let q  Z(r, f − αj ) + N (r, f ) − Z0Y (r, f  ) − log r + f ∈ M(IK) (resp. f ∈ Mu (S)). Then (q − 1)T (r, f ) ≤ j=1

O(1)

∀r ∈ I (resp. ∀r ∈ J).

Moreover, if f belongs to A(IK) (resp. if f ∈ A(S)), then (q − 1)T (r, f ) ≤

q 

Z(r, f − αj ) −

j=1

Z0Y (r, f  ) − log(r) + O(1) ∀r ∈ I (resp. ∀r ∈ J). 3. NEVANLINNA THEORY OUT OF A HOLE Henceforth, we denote by L the interval [R, +∞[. According to classical properties of analytic elements on infraconnected sets and by implicitly using [8], it is easy to have the following properties: q  Proposition 3.1. Let f ∈ H(D) have no zero in D. Then f (x) is of the form an xn with −∞

|aq |Rq > |an |Rn for all n < q. Definition and notation. Let f ∈ H(D) have no zero in D, f (x) =

q 

an xn with |aq |Rq > |an |Rn for

−∞

all n < q and aq = 1. Then f will be called a Motzkin factor associated to S and the integer q will be called the Motzkin index of f and will be denoted by m(f, S) (see [2, 8, 15]). Theorem 3.2. Let f ∈ M(D). We can write f in a unique way in the form f S f 0 with f S ∈ H(D) a Motzkin factor associated to S and f 0 ∈ M(IK), having no zero and no pole in S. Given f ∈ M(D), for r > R. If α1 , ..., αm are the distinct zeros of f in Δ(0, R, r), with respective multiplicity uj , 1 ≤ j ≤ m, then the counting function of zeros ZR (r, f ) of f between R and r will be denoted by m  uj (log(r) − log(|αj |)). Similarly, if β1 , ..., βn are the distinct poles of f in Δ(0, R, r), ZR (r, f ) = j=1

with respective multiplicity vj , 1 ≤ j ≤ m, then the counting function of poles NR (r, f ) of f between R and r will be denoted by n    vj (log(r) − log(|βj |)). We put TR (r, f ) = max ZR (r, f ), NR (r, f ) . NR (r, f ) = j=1

The counting function of zeros without counting multiplicity Z R (r, f ) is defined as: Z R (r, f ) = m  log(r) − log(|αj |), where α1 , ..., αm are the distinct zeros of f in Δ(0, R, r). Similarly, the counting j=1

function of poles without counting multiplicity N R (r, f ) is defined as: n  N R (r, f ) = log(r) − log(|βj |), where β1 , ..., βn are the distinct poles of f in Δ(0, R, r). j=1 Y (r, f  ) the counting function of zeros of f  on points Finally, putting Y = {a1 , ..., aq }, we denote by ZR x where f (x) ∈ / Y.

Theorem 3.3. Let f ∈ M(D). Then, for all r ∈ L, log(|f |(r)) − log(|f |(R)) = ZR (r, f ) − NR (r, f ) + m(f, S)(log r − log R).

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Corollary 3.3.a. Let f ∈ M(D). Then TR (r, f ) is identically zero if and only if f is a Motzkin factor. Let f, g ∈ A(D) satisfy log(|f |(r)) ≤ log(|g|(r)) for all r ∈ L. Then ZR (r, f ) ≤ ZR (r, g) + (m(g, S) − m(f, S))(log(r) − log(R)). Theorem 3.4. Let f ∈ A(D). Then, ZR (r, f  ) ≤ ZR (r, f ) − log(r) + O(1), (r ∈ L). We can now characterize the set M∗ (D): Theorem 3.5. Let f ∈ M(D). The three following statements are equivalent: TR (r, f ) = +∞ for r ∈ L, r→+∞ log(r)

i) lim ii)

TR (r, f ) is unbounded, log(r)

iii) f belongs to M∗ (D). Operations on M(D) work almost like for meromorphic functions in the whole field [4, 8]. Theorem 3.6. If f, g ∈ M(D). Then for every b ∈ IK and r ∈ L, we have TR (r, f n ) = nTR (r, f ), TR (r, f.g) ≤ TR (r, f ) + TR (r, g) + O(log(r)), 1 TR (r, ) = TR (r, f )), TR (r, f + g) ≤ TR (r, f ) + TR (r, g) + O(log(r)), f TR (r, f + b) = TR (r, f ) + O(log(r)), TR (r, h ◦ f ) = TR (r, f ) + O(log(r)), where h is a Moebius function. Moreover, if both f and g belong to A(D), then TR (r, f + g) ≤ max(TR (r, f ), TR (r, g)) + O(log(r)), TR (r, f g) = TR (r, f ) + TR (r, g). Particularly, if f ∈ A∗ (D), then TR (r, f + b) = TR (r, f ) + O(1). Given a polynomial P (x) ∈ IK[x], then TR (r, P ◦ f ) = deg(P )TR (r, f ) + O(log(r)). Theorem 3.7. Every f ∈ M∗ (D) is transcendental over Mz (D). Theorem 3.8. Let f ∈ M(D). Then, for r ∈ L, NR (r, f (k) ) = NR (r, f ) + kN R (r, f ) and ZR (r, f (k) ) ≤ ZR (r, f ) + kN R (r, f ) + O(log(r)). Like in the whole field, the Nevanlinna second Main Theorem is based on the following theorem: Theorem 3.9. Let f ∈ M(D)and let a1 , ..., aq ∈ IK bedistinct. Then q (q − 1)TR (r, f ) ≤ max1≤k≤q j=1,j=k ZR (r, f − aj ) + O(log(r)). Theorem 3.10: (Second Main Theorem). Let f ∈ M(D), let α1 , ..., αq ∈ IK, with q ≥ 2 and let Y = {α1 , ..., αq }. Then, for r ∈ L, (q − 1)TR (r, f ) ≤

q 

Y Z R (r, f − αj ) + N R (r, f ) − ZR (r, f  ) + O(log(r)).

j=1

Particularly, if f ∈ A(D), then (q − 1)TR (r, f ) ≤

q 

Y Z R (r, f − αj ) − ZR (r, f  ) + O(log(r)).

j=1

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4. IMMEDIATE APPLICATIONS Theorem 4.1. Let a1 , a2 ∈ IK (with a1 = a2 ) and let f, g ∈ A∗ (IK) satisfy f −1 ({ai }) = g−1 ({ai }) (i = 1, 2). Then f = g. Theorem 4.2. Let a1 , a2 , a3 ∈ IK (with ai = aj ∀i = j) and let f, g ∈ Au (d(0, R− )) (resp. f, g ∈ A∗ (D)) satisfy f −1 ({ai }) = g−1 ({ai }) (i = 1, 2, 3). Then f = g. Theorem 4.3. Let a1 , a2 , a3 , a4 ∈ IK (with ai = aj ∀i = j) and let f, g ∈ M∗ (IK) satisfy f −1 ({ai }) = g−1 ({ai }) (i = 1, 2, 3, 4). Then f = g. Theorem 4.4. Let a1 , a2 , a3 , a4 , a5 ∈ IK (with ai = aj ∀i = j) and let f, g ∈ Mu (d(0, R− )) (resp. f, g ∈ M∗ (D)) satisfy f −1 ({ai }) = g−1 ({ai }) (i = 1, 2, 3, 4, 5). Then f = g. Theorem 4.5. Let Λ be a non-degenerate elliptic curve of equation y 2 = (x − a1 )(x − a2 )(x − a3 ). There do not exist g, f ∈ M(IK) such that g(t) = y, f (t) = x, t ∈ IK. There do not exist g, f ∈ Au (d(0, R− )) such that g(t) = y, f (t) = x, t ∈ d(0, R− ). There do not exist g, f ∈ A∗ (D) such that g(t) = y, f (t) = x, t ∈ D. Theorem 4.6. Let Λ be a curve of equation y q = P (x), q ≥ 2, with P ∈ IK[x] admitting n distinct zeros of order 1 with n ≥ 4. There do not exist g, f ∈ M(IK) such that g(t) = y, f (t) = x, t ∈ IK. Theorem 4.7. Let Λ be a curve of equation y q = P (x), q ≥ 2, with P ∈ IK[x] admitting n distinct zeros of order 1 with n ≥ 5. There do not exist g, f ∈ Mu (d(0, R− )) (resp. g, f ∈ M∗ (D)) such that g(t) = y, f (t) = x, t ∈ d(0, R− ) (resp. t ∈ D). Another application concerns analytic functions: Theorem 4.8. Let f, g ∈ M(d(0, R− )) (resp. f, g ∈ M(D)) satisfy gm + f n = 1, with min(m, n) ≥ 3 and max(m, n) ≥ 4). Then f and g belong to Mb (d(0, R− )) (resp. to Mz (D)). Moreover, if f, g ∈ A(d(0, R− )) (resp. if f, g ∈ A(D)) satisfy gm + f n = 1, with min(m, n) ≥ 2 and (m, n) = (2, 2), then f and g belong to Ab (d(0, R− )), (resp to Az (D)). 5. SMALL FUNCTIONS Definitions. For each f ∈ M(IK), (resp. f ∈ M(S), resp. f ∈ M(D)), we will denote by Mf (IK) (resp. Mf (S), Mf (D)) the set of functions h ∈ M(IK) (resp. h ∈ M(S), h ∈ M(D)) such that T (r, h) = o(T (r, f )), r ∈ I (resp. T (r, h) = o(T (r, f )), r ∈ J, resp. TR (r, h) = o(TR (r, f )), r ∈ L). Similarly, if f ∈ A(IK) (resp. f ∈ A(S), resp. f ∈ A(D)) we will denote by Af (IK) (resp. Af (S), resp. Af (D)) the set Mf (IK) ∩ A(IK),( resp. Mf (S) ∩ A(S), resp. Mf (D) ∩ A(D)). The elements of Mf (IK) (resp. Mf (S), resp. Mf (D)) are called small meromorphic functions with respect to f . Similarly, if f ∈ A(IK) (resp. f ∈ A(S), resp. f ∈ A(D)) these functions are called small analytic functions with respect to f . A small function w with respect to a function f ∈ M(IK) (resp. f ∈ M(S), resp. f ∈ M(D)) will be called a quasi-exceptional small function for f if f − w has finitely many zeros in D. Theorem 5.1. Let f ∈ M∗ (IK) (resp. f ∈ Mu (S), resp. f ∈ M∗ (D)). Then f admits at most one quasi-exceptional small function. Moreover, if f has finitely many poles, then f admits no quasiexceptional small function. Corollary 5.1.a. Let f ∈ A∗ (IK) (resp. f ∈ Au (S), resp. f ∈ A∗ (D)). Then f has no quasiexceptional small function. The Second Main Theorem for Three Small Functions holds as well as in complex analysis. Notice that this theorem was generalized to any finite set of small functions by K. Yamanoi in complex analysis [17], through methods that have no equivalent on a p-adic field. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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Theorem 5.2. Let f ∈ M∗ (IK), (resp. f ∈ Mu (S), resp. f ∈ M∗ (D)) and let w1 , w2 , w3 ∈ Mf (IK)  (resp. ∈ Mf (d(0, R− )), resp. ∈ Mf (D)) be pairwise distinct. Then: T (r, f ) ≤ 3j=1 Z(r, f − wj ) + o(T (r, f ))   (resp. T (r, f ) ≤ 3j=1 Z(r, f − wj ) + o(T (r, f )), resp. TR (r, f ) ≤ 3j=1 Z R (r, f − wj ) + o(TR (r, f ))). Corollary 5.2.a. Let f ∈ M∗ (IK) (resp. f ∈ Mu (S) resp. f ∈ M∗ (D)) and let w1 , w2 ∈ Af (IK) (resp. w1 , w2 ∈ Af (S) resp. w1 , w2 ∈ Af (D)) be distinct. Then T (r, f ) ≤ Z(r, f − w1 ) + Z(r, f − w2 ) + N (r, f ) + o(T (r, f )), (resp. T (r, f ) ≤ Z(r, f − w1 ) + Z(r, f − w2 ) + N (r, f ) + o(TR (r, f )), resp. TR (r, f ) ≤ Z R (r, f − w1 ) + Z R (r, f − w2 ) + N R (r, f ) + o(TR (r, f ))). Corollary 5.2.b. Let f ∈ A∗ (IK) (resp. f ∈ Au (S), resp. f ∈ A∗ (D)) and let w1 , w2 ∈ Af (S), resp. w1 , w2 ∈ Af (D)) be distinct. Then T (r, f ) ≤ Z(r, f − w1 ) + Z(r, f − w2 ) + o(T (r, f )), (resp. T (r, f ) ≤ Z(r, f − w1 ) + Z(r, f − w2 ) + o(T (r, f )), resp. TR (r, f ) ≤ Z R (r, f − w1 ) + Z R (r, f − w2 ) + o(TR (r, f ))). Definitions. Let f ∈ M∗ (IK) (resp. f ∈ Mu (d(0, R− )), resp. f ∈ M∗ (D)) and let w ∈ Mf (IK) (resp. w ∈ Mf (d(0, R− )), resp. w ∈ Mf (D)). Then w is called a perfectly branched function with respect to f if all zeros of f − w are multiple except maybe finitely many [3], [8] and w is called a totally branched function with respect to f if all zeros of f − w are multiple, without exception [6]. Particularly, the definition applies to constants. The following Theorem 5.3, 5.4, 5.5 are proved in [8] concerning the sets f ∈ M∗ (IK) (resp. f ∈ Mu (d(0, R− )), f ∈ A∗ (IK) (resp. f ∈ Au (d(0, R− )). The proofs concerning M∗ (D) and A∗ (D) are similar to those concerning M∗ (IK) and A∗ (IK) respectively. Theorem 5.3. Let f ∈ M∗ (IK) (resp. f ∈ Mu (d(0, R− )), resp. f ∈ M∗ (D)). Then f admits at most four perfectly branched values. Theorem 5.4. Let f ∈ M∗ (IK) (resp. f ∈ M∗ (D)) having finitely many poles. Then f admits at most one perfectly branched rational function. Corollary 5.4.a. Let f ∈ A∗ (IK) (resp. f ∈ A∗ (D)). Then f admits at most one perfectly branched rational function. Theorem 5.5. Let f ∈ Mu (d(0, R− )), having finitely many poles. Then f admits at most two perfectly branched rational functions. Corollary 5.5.a. Let f ∈ Au (d(0, R− )). Then f admits at most two perfectly branched rational functions. 6. NEW APPLICATIONS OF THE NEVANLINNA THEORY Definitions. Recall that two functions f and g meromorphic in a set B are said to share a set X ⊂ IK, counting multiplicity, or C.M. in brief, if for each b ∈ X, when f (x) − b has a zero of order q at a point a ∈ B, then there exists c ∈ X such that g(x) − c also has a zero of order q at a. And the functions f and g are said to share X ignoring multiplicity or I.M. in brief, if f −1 (X) = g−1 (X). In the same way, two functions f and g meromorphic in a set B are said to share a nonidentically g f zero function h from B to IK, counting multiplicity, or C.M. in brief, if and share 1 C.M., i.e. if h h f − h and g − h have the same zeros with the same multiplicity. By Theorem 4.3, two meromorphic functions in IK sharing 4 points I.M. are identical , by Theorem 4.4 two meromorphic functions in S or in D sharing 5 points I.M. are identical, by Theorem 4.1, two entire functions sharing 2 points I.M. are identical and by Theorem 4.2 two meromorphic functions in S p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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or in D sharing 3 points I.M. are identical. Here we will first examine two meromorphic functions sharing a few points C.M. Theorem 6.1. Let f, g ∈ Mu (S) (resp. let f, g ∈ M∗ (D)) share C.M. 4 points aj ∈ IK ∪ {∞}, j = 1, 2, 3, 4. Then f ≡ g. Theorem 6.2. Let f, g ∈ Mu (S) (resp. let f, g ∈ M∗ (D)) have finitely many poles and share C.M. 3 points aj ∈ IK ∪ {∞}, j = 1, 2, 3. Then f ≡ g. Corollary 6.2.a. Let f, g ∈ Au (S) (resp. f, g ∈ A∗ (D)) share C.M. 3 points aj ∈ IK, Then f ≡ g.

j = 1, 2, 3.

Theorem 6.3 is not immediate and has a similar version in complex analysis for meromorphic functions provided with a finite growth order that is not integral. Here, we don’t need any hypothesis on the growth order. Theorem 6.3. Let f, g ∈ M∗ (IK) share C.M. 3 points aj ∈ IK ∪ {∞}, j = 1, 2, 3. Then f ≡ g. In the particular case of functions f, g ∈ Mu (S) or functions f, g ∈ M∗ (D)) having finitely many poles and sharing poles C.M., we can add this theorem: Theorem 6.4. Let f, g ∈ Mu (S), (resp. let f, g ∈ M∗ (D)) have finitely many poles in S (resp. in D) and share C.M. two values a, b and poles. Then f ≡ g. Our main theorems are Theorems 6.5 and 6.6 that follow the same kind of reasoning as in [7]. We denote by Y1 = {α1 , ..., αk } and Y2 = {β1 , β2 } the two sets satisfying (H)

k 

k 2   2 (β1 − αj ) = (β2 − αj ) .

j=1

j=1

Theorem 6.5. Let f, g ∈ M∗ (IK) (resp. let f, g ∈ Mu (S), resp. let f, g ∈ M∗ (D)) have finitely many poles in IK (resp. in S, resp. in D) and share Y1 C.M. and Y2 I.M. Then f ≡ g. Corollary 6.5.a. Let f, g ∈ M∗ (IK) (resp. let f, g ∈ Mu (S), resp. let f, g ∈ M∗ (D)) have finitely many poles in IK (resp. in S, resp in D) and share a value α C.M. and Y2 I.M. If (α − β1 )2 = (α − β2 )2 , then f ≡ g. Corollary 6.5.b. Let f, g ∈ A∗ (IK) (resp. let f, g ∈ Au (S), resp. let f, g ∈ A∗ (D)) and share Y1 C.M. and Y2 I.M. Then f ≡ g. Corollary 6.5.c. Let f, g ∈ A∗ (IK) (resp. let f, g ∈ Au (S), resp. let f, g ∈ A∗ (D)) and share a value α C.M. and Y2 I.M. If (α − β1 )2 = (α − β2 )2 , then f ≡ g. Theorem 6.6. Let f, g ∈ M∗ (IK) (resp. let f, g ∈ Mu (S), resp. let f, g ∈ M∗ (D)) have finitely many poles in IK (resp. in S, resp. in D) and share Y1 I.M. and Y2 C.M. Then f ≡ g. Corollary 6.6.a. Let f, g ∈ M∗ (IK) (resp. let f, g ∈ Mu (S), resp. let f, g ∈ M∗ (D)) have finitely many poles in IK (resp. in S, resp. in D) and share a value α I.M. and Y2 C.M. If (α − β1 )2 = (α − β2 )2 , then f ≡ g. Corollary 6.6.b. Let f, g ∈ A∗ (IK) (resp. let f, g ∈ Au (S), resp. let f, g ∈ A∗ (D)) and share Y1 I.M. and Y2 C.M. Then f ≡ g. Corollary 6.6.c. Let f, g ∈ A∗ (IK) (resp. let f, g ∈ Au (S), resp. let f, g ∈ A∗ (D)) and share a value α I.M. and Y2 C.M. If (α − β1 )2 = (α − β2 )2 , then f ≡ g. It is known that if two functions f, g ∈ A(IK) share separately two values a, b ∈ IK C.M., then f ≡ g [8] (Theorem 41.1). However, here the hypothesis f, g share Y1 and share Y2 cannot be compared: for example, concerning Y2 , f and g are not supposed to share β1 or β2 separately. The same remark applies to meromorphic functions having finitely many poles. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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Results of [16] showed the interest of complex functions of the form f (x)f (x + b). Similar studies were made in [12] and [14] in a p-adic field. Here we will generalize that kind of study on the field IK. In [16], it is proven that if two complex entire functions f and g are such that f (x)n f (x + c) and f g(x)n g(x + c) share 1 C.M. with n ≥ 6, then either f g is a constant t1 such that tn+1 = 1, or is a 1 g = 1. Here, on the field IK, we can obtain better results. constant t2 such that tn+1 2 On the other hand, in [12], we can find similar results of uniqueness for meromorphic functions on a p-adic field involving derivatives, sharing 1 C.M. or I.M., also involving derivatives. Here we will examine functions of the form f (x)n (f (x + c))m , g(x)n (g(x + c))m sharing a rational function and we will look for branched values and quasi-exceptional values of such functions. Notation. We denote by IN∗ the set of strictly positive integers. On ZZ, we denote by | . |∞ the Archimedean absolute value. Theorem 6.7. Let a ∈ C(0, 1), let b ∈ IK and let f, g ∈ M∗ (IK) have finitely many poles and take m, n IN∗ with m = n. If f n (x)f m (ax + b) and gn (x)gm (ax + b) share C.M. a rational function f Q ∈ IK(x), Q ≡ 0 and if n + m ≥ 5, then is a constant t such that tn+m = 1. Moreover, if f, g ∈ g A∗ (IK), if f n (x)f m (ax + b) and gn (x)gm (ax + b) share C.M. a constant l = 0 and if n + m ≥ 4, then f is a constant t such that tn+m = 1. g Theorem 6.8. Let a ∈ C(0, 1) and let b ∈ S and let f, g ∈ Mu (S) (resp. let f, g ∈ M∗ (D)) have finitely many poles in S (resp. in D) and take n, m ∈ IN∗ with n = m. If f n (x)f m (ax + b) and g n (x)gm (ax + b) share C.M. a function θ ∈ M(S) (resp. a function θ ∈ M(D)) having finitely many f zeros and poles in S (resp. in D), if n ≥ 5, then is a constant t such that tn+m = 1. g In [16] it was shown that given a complex entire function f and b ∈ C l \ {0}, a function of the form l provided n ≥ 3. On the field IK, such a f n (x)f (x + b) − c (with c = 0) has infinitely many zeros in C result is trivial since an entire functions and even a meromorphic function with finitely many poles (which is not a rational function) takes every value infinitely many times. But we can ask the question regarding in general functions f ∈ M∗ (IK), f ∈ Mu (S), f ∈ M∗ (D). Theorem 6.9. Let f ∈ M∗ (IK) (resp. f ∈ Mu (S), resp. f ∈ M∗ (D)), let a ∈ C(0, 1), let b ∈ IK (resp. b ∈ S, resp. b ∈ S) and let w ∈ M(IK) (resp. w ∈ M(S), resp. w ∈ M(D)) be a non identically zero small function with respect to f . If n, m ∈ IN are such that |n − m|∞ ≥ 5, then f n (x)f m (ax + b) − w has infinitely many zeros in IK (resp. in S, resp. in D). Corollary 6.9.a. Let f ∈ M∗ (IK) (resp. f ∈ Mu (S), resp. f ∈ M∗ (D)), let a ∈ C(0, 1), let b ∈ IK (resp. b ∈ S, resp. b ∈ S). If n, m ∈ IN are such that |n − m|∞ ≥ 5, then f n (x)f m (ax + b) takes every nonzero value infinitely many times in IK (resp. in S, resp. in D). Remarks. 1) Of course, the hypothesis w = 0 must not be excluded. Indeed, let h ∈ A∗ (IK) and let 1 . Then a function of the form f n (x)f (ax + b) has no zero in IK. f (x) = h(x) 2) On the other hand, it is known and easily seen that if a = 1 and b = 0, a function f n − w has infinitely many zeros for every n ≥ 3 and that f 2 takes every nonzero value c infinitely many times because given a square root l of c, then f 2 − c = (f − l)(f + l) and at least one of the two values l and −l is taken infinitely many times. We can ask whether a meromorphic function of the form f 2 (x) − w always has infinitely many zeros when w is not a constant. 3) Concerning Theorem 6.9, it is easily proven that if f is a meromorphic function with finitely many poles and w a small function, then f − w has infinitely many zeros. So it is useless here to add a corollary concerning f n (x)f m (ax + b) − w when f has finitely many poles. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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Theorem 6.10. Let f ∈ M∗ (IK) (resp. f ∈ Mu (S), resp. f ∈ M∗ (D)), let a ∈ C(0, 1), let b ∈ IK (resp. b ∈ S, resp. b ∈ S) and let n, m ∈ IN∗ . If 3|n − m|∞ > 2(n + m) + 4, then f n (x)f m (ax + b) does not admit 4 distinct perfectly branched values. Moreover, if 4|n − m|∞ > 3(n + m) + 4, then f n (x)f m (ax + b) does not admit 3 distinct perfectly branched values. Further, if f belongs to M∗ (IK) and if 3|n − m|∞ ≥ 2(n + m + 1), then f n (x)f m (ax + b) does not admit 4 distinct totally branched values; moreover, if 4|n − m|∞ ≥ 3(n + m) + 4, then f n (x)f m (ax + b) does not admit 3 distinct totally branched values. 7. TOOLS AND LEMMAS In the proofs of our main theorems, we will use the following lemmas: Lemma 1. Let P (x) ∈ IK[x] have all its zeros of order 1. Let f ∈ M∗ (IK) (resp. let f ∈ Mu (S), resp. let f ∈ M∗ (D)) and suppose that f − b admits a zero of order q at a point u ∈ IK (resp. at a point u ∈ S, resp. at a point u ∈ D). Then P (f ) − P (b) admits u as a zero of order q. Lemmas 2, 3 and 4 are well known [8], Lemma 32.3, Corollary 32.6. Lemma 2. Let f ∈ M∗ (IK) have finitely many poles. Then for every Q ∈ IK(x), f − Q has inifinitely many zeros. Lemma 3. Let f ∈ M(IK). There exist φ and ψ ∈ A(IK) having no common zero, such that f =

φ . ψ

Moreover, if f has no zero and no pole then f is a constant. Lemma 4 is a consequence of Theorem 40.10 and Theorem 43.1 in [8]: Lemma 4. Let w ∈ M(S) have finitely many zeros and poles in S. Then w belongs to Mb (S). Moreover, given f ∈ Mu (S) having finitely many poles in S, then f − w belongs to Mu (S) and has infinitely many zeros. Lemma 5. Let f ∈ A(D). Then f belongs to Az (D) if and only if it is of the form

q 

an xn with

−∞

q ∈ ZZ and lim |an |Rn = 0. Moreover, given f ∈ M∗ (D) having finitely many poles in D and n→−∞

w ∈ Mz (D), then f − w has infinitely many zeros. Lemma 6 is immediate : Lemma 6. Let u ∈ C(0, 1) and c ∈ IK (resp. c ∈ S, resp. c ∈ S) and let f ∈ M(IK) (resp. f ∈ M(S), resp. f ∈ M(D)). Then T (r, f ) = T (r, f (ux + c)), Z(r, f ) = Z(r, f (ux + c)), N (r, f ) =N (r, f (ux + c)) whenever r ≥ |c| (resp. T (r, f ) = T (r, f (ux + c)), Z(r, f ) = Z(r, f (ux + c)), N (r, f ) = N (r, f (ux +c)) whenever r ≥ |c|, resp. TR (r, f ) = TR (r, f (ux + c)) ZR (r, f ) = ZR (r, f (ux + c)), NR (r, f ) = NR (r, f (ux + c)) whenever r ≥ |c|). Notation. Let f, g ∈ M(IK) (resp. let f, g ∈ M(S), resp. let f, g ∈ M(D)). We set M (r, f ) = max((Z(r, f ), N (r, f )) − N (r, f ) (resp. M (r, f ) = max((Z(r, f ), N (r, f )) − N (r, f ), resp. MR (r, f ) = max((ZR (r, f ), NR (r, f )) − NR (r, f )). Lemma 7 is also easily proven by the property M (r, f ) = max(log(|f |(r)), 0) on M(IK), and on M(S) and MR (r, f ) = max(log(|f |(r)), 0) on M(D). Lemma 7. Let f, g ∈ M(IK) (resp. let f, g ∈ M(S)). Then M (r, f g) ≤ M (r, f ) + M (r, g), M (r, f + f g) ≤ M (r, f ) + M (r, g), M (r, ) ≤ O(1). f Let f, g ∈ M(D). Then MR (r, f g) ≤ MR (r, f ) + MR (r, g), MR (r, f + g) ≤ MR (r, f ) + MR (r, g), f MR (r, ) ≤ O(1). f p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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Lemmas 8 and 9 are easily proven: Lemma 8. Let f, g ∈ M∗ (IK) (resp. let f, g ∈ Mu (S), resp. let f, g ∈ M∗ (D)) have finitely many poles in IK (resp. in S, resp. in D). Then T (r, λf + μg) ≤ max(T (r, f ), T (r, g)) + O(1) (resp. T (r, λf + μg) ≤ max(T (r, f ), T (r, g)) + O(1) , resp. TR (r, λf + μg) ≤ max(TR (r, f ), TR (r, g)) + O(log(r))). Lemma 9. Let f, g ∈ M∗ (IK) (resp. let f, g ∈ Mu (S), resp. let f, g ∈ M∗ (D)) have finitely 1 1 many poles in IK (resp. in S, resp. in D). Then Z(r, − ) ≤ Z(r, f − g) + O(log(r))), (resp. f g 1 1 1 1 Z(r, − ) ≤ Z(r, f − g) + O(1), resp. ZR (r, − ) ≤ ZR (r, f − g) + O(log(r)))). f g f g 8. PROOF OF THEOREMS Proof of Theorems 6.1 and 6.2. Suppose f = g. We will first prove Theorems 6.1 when f and g belong to Mu (S). So, let f, g ∈ Mu (S) sharing C.M. q points aj ∈ IK, 1 ≤ j ≤ q. By Theorem 2.7, we have (q − 1)T (r, f ) ≤

q 

Z(r, f − aj ) + O(1) ≤ Z(r, f − g) + O(1))

j=1

≤ T (r, f ) + T (r, g) + O(1)

(1) and similarly (q − 1)T (r, g) ≤

q 

Z(r, g − aj ) + O(1) ≤ Z(r, f − g) + O(1)

j=1

≤ T (r, f ) + T (r, g) + O(1)

(2) hence

(q − 1)(T (r, f ) + T (r, g)) ≤ 2(T (r, f ) + T (r, g)) + O(1),

(3)

hence (3) implies q ≤ 3. Suppose now that f and g have finitely many poles. If f and g belong to Mu (S), by Lemma 8, we have Z(r, f − g) ≤ max(T (r, f ), T (r, g)), hence by (1) and (2), now we obtain (q − 1) max(T (r, f ), T (r, g)) ≤ max(T (r, f ), T (r, g)) + O(1) and hence q ≤ 2. Consequently, since q = 3, we have f ≡ g in Theorem 6.2. If f, g ∈ M∗ (D), the same reasoning holds. Proof of Theorem 6.3. Let f, g ∈ M∗ (IK) be distinct and share three points aj ∈ IK, j = 1, 2, 3 C.M. Without loss of generality, we can assume that a1 a2 = 0. Suppose first that f (x) = a3 ∀x ∈ IK, hence 1 1 1 j = 1, 2. Then F and G belong to to A(IK) and g(x) = a3 ∀x ∈ IK. Let F = , G = and let bj = f g aj share b1 and b2 C.M and hence, by Theorem 4.1 we have F ≡ G, hence f ≡ g. Thus, henceforth, we can assume that f and g take each value aj , j = 1, 2, 3 at least one time, therefore Z(r, f − aj ) ≥ log(r) + O(1), r ∈]0, +∞[, j = 1, 2, 3.

(1) Now we notice that (2)

3 

Z(r, f − aj ) ≤ Z(r, f − g) ≤ T (r, f ) + T (r, g).

j=1

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Suppose first that lim

r→+∞

3 4

23

Z(r, f − aj ) = 1 ∀j = 1, 2, 3. Then there exists s > 0 such that T (r, f )

Z(r,f −aj ) T (r,f )

≥

∀j = 1, 2, 3, ∀r > s. Then for every r > s, we have  9 T (r, f ) ≤ Z(r, f − aj ) ≤ Z(r, f − g) + O(1) ≤ T (r, f ) + T (r, g) + O(1) 4 3

j=1

and similarly  9 T (r, g) ≤ Z(r, g − aj ) ≤ Z(r, f − g) + O(1) ≤ T (r, f ) + T (r, g) + O(1). 4 3

j=1

9 Therefore, we obtain (T (r, f ) + T (r, g)) ≤ 2(T (r, f ) + T (r, g)) + O(1), a contradiction showing that 4 Z(r, f − aj ) < 1 holds for at least one index. lim inf r→+∞ T (r, f ) Consequently, we are now led to assume there exists  > 0 such that for a certain k, there exists a sequence of intervals ([sn , tn ])n∈ IN such that sn < tn < sn+1 , limn→+∞ sn = +∞ and Z(r, f − ak ) ≤ 1 −  ∀r ∈ [sn , tn ], ∀n ∈ IN. Without loss of generality, we can suppose k = 3 and set T (r, f ) +∞ [sn , tn ]. So, when r belongs to B, we have B= n=1

Z(r, f − a3 ) ≤ (1 − )T (r, f ) ∀r ∈ B.

(3)

φ with φ, ψ ∈ A(IK) having no common zero. Then N (r, f − ψ a3 ) = N (r, f ) = Z(r, ψ), hence by (3), we have

By Lemma 3 we can write f in the form

T (r, f − a3 ) = T (r, f ) = N (r, f ) ∀r ∈ B. Now Z(r, f − aj ) = Z(r, φ − aj ψ), j = 1, 2, 3, N (r, f ) = Z(r, ψ). Therefore by (3), we have Z(r, φ − a3 ψ) ≤ (1 − )Z(r, ψ).

(4)

By (4) and by Theorem 2.1, we can derive log(|φ − a3 ψ|(r)) ≤ (1 − ) log(|(a3 − a1 )ψ|(r)) + O(1) and hence log(|φ − a3 ψ|(r)) < log(|(a3 − a1 )ψ|(r)) when r is big enough in B. So, there exists u ≥ s such that log(|φ − a3 ψ|(r)) < log(|(a3 − a1 )ψ|(r)) ∀r ∈ B, r ≥ u. Consequently, log(|φ − a1 ψ|(r)) = log |(a3 − a1 )ψ|(r) ∀r ∈ J, r ≥ u and hence Z(r, f − a1 ) ≥ Z(r, ψ) + O(1) ∀r ∈ B, r ≥ u, therefore Z(r, f − a1 ) = T (r, f ) + O(1) ∀r ∈ B, r ≥ u.

(5) Similarly, we have

Z(r, f − a2 ) = T (r, f ) + O(1) ∀r ∈ B, r ≥ u.

(6)

Consequently, by (5) and (6), we obtain Z(r, f − a1 ) + Z(r, f − a2 ) ≥ 2T (r, f ) + O(1) r ∈ J, r ≥ u and  hence 3j=1 Z(r, f − aj ) ≥ 2T (r, f ) + Z(r, f − a3 ) + O(1) r ∈ B, r ≥ u, therefore by (2) 2T (r, f ) + Z(r, f − a3 ) ≤ T (r, f ) + T (r, g) + O(1) r ∈ B, r ≥ u and similarly 2T (r, g) + Z(r, g − a3 ) ≤ T (r, f ) + T (r, g) + O(1) r ∈ B, r ≥ u. Thus, 2(T (r, f ) + T (r, g)) + Z(r, f − a3 ) + Z(r, g − a3 ) ≤ 2(T (r, f ) + T (r, g)) + O(1), r ∈ B, r ≥ u. But then, by (1) we have a contradiction that finishes the proof. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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Proof of Theorem 6.4. Suppose f = g. Without loss of generality, we can assume ab = 0. Let 1 1 1 1 F = , G= and let a = , b = . Then F and G share 0, a , b C.M. Suppose that F f g a b and G are not identical. By Theorems 2.7 and 3.10, we have 2T (r, F ) ≤ Z(r, F ) + Z(r, F − a ) + Z(r, F − b ) + O(1) ≤ Z(r, F − G) + O(1) (resp. 2TR (r, F ) ≤ ZR (r, F ) + ZR (r, F − a ) + ZR (r, F − b ) + O(log(r)) ≤ ZR (r, F − G) + O(log(r))) hence by Lemma 9, 2T (r, F ) ≤ T (r, f − g) + O(1) (resp. 2TR (r, F ) ≤ TR (f − g)) and similarly, 2T (r, G) ≤ T (r, f − g) + O(1), (resp. 2TR (r, G) ≤ TR (r, f − g) + O(log(r))). Consequently, by Theorem 2.2 and 3.6 and Lemma 8, we have 2 max(T (r, f ), T (r, g)) ≤ max(T (r, f ), T (r, g)) + O(1) (resp. 2 max(TR (r, f ), TR (r, g) ≤ max(TR (r, f ), TR (r, g) + O(log(r))), a contradiction since f, g belong to Mu (S), (resp. to M∗ (D)). Consequently, F ≡ G, and hence f ≡ g, a contradiction. Proof of Theorem 6.5. Let H ∈ IK(x) (resp. let H ∈ Mb (S), resp. H ∈ Mz (D)) be such that the

H(x) kj=1 (f (x) − αj ) have no pole and no zero in IK (resp. in S, resp. in D). Then by function V (x) =

k j=1 (g(x) − αj ) Lemma 3, V is a constant in IK (resp. V belongs to Mb (S), resp. V belongs to Mz (D)). On the other hand, since f and g have finitely many poles, in all hypotheses, we note that N (r, f ) = O(log(r)), N (r, g) = O(log(r)). Since f and g share Y2 I.M., we can derive that each zero of (f (x) − β1 )(f (x) − β2 ) in IK (resp. in S, resp. in D) is a zero of one of the following three functions: H −1 V − 1,

H −1 V −

k  β1 − αj , β2 − αj

H −1 V −

j=1

k  β2 − αj . β1 − αj

j=1

We can then derive that one of the following three cases a), b), c) must occur:

a)

k  f (x) − αj ≡1 g(x) − αj

j=1

b)

k k   f (x) − αj β1 − αj ≡ g(x) − αj β2 − αj

j=1

c)

j=1

k k   f (x) − αj β2 − αj ≡ . g(x) − αj β1 − αj

j=1

j=1

Indeed, suppose that none of the cases a), b), c) occurs. Then the following three inequalities are satisfied:

(i)

k  f (x) − αj ( ) ≡ 1 g(x) − αj

j=1

(ii)

k k   f (x) − αj β1 − αj ( ) ≡ ( ) g(x) − αj β2 − αj

j=1

(iii)

j=1

k k   f (x) − αj β2 − αj ( ) ≡ ( ). g(x) − αj β1 − αj

j=1

j=1

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Suppose first, for simplicity, that f and g belong to M∗ (IK). Then, by Theorem 2.7, we have: T (r, f ) ≤ Z(r, f − β1 ) + Z(r, f − β2 ) + N (r, f ) + O(log(r)) k k     β1 − αj  β2 − αj  ( ) + Z r, H −1 V − ( ) + O(log(r)) ≤ Z(r, H −1 V − 1) + Z r, H −1 V − β2 − αj β1 − αj j=1

j=1

≤ 3T (r, H −1 V ) + O(log(r)) ≤ O(log(r)). Consequently, T (r, f ) = O(log(r)) i.e. f ∈ IK(x), a contradiction. Similarly, if f and g belong to Mu (S) or to M∗ (D), following the same calculation, we obtain f ∈ Mb (S), f ∈ Mz (D), the same contradiction. Therefore one of the cases a), b), c) must occur. We will examine them successively. Suppose a) is satisfied. Since f and g share Y2 I.M., thanks to Hypothsis (H), we can see that f (x) = β1 if and only if g(x) = β1 and f (x) = β2 if and only if g(x) = β2 . Consequently, f and g share both β1 and β2 I.M. Then, by Lemma 1 and Relation a), this implies that f and g share β1 C.M. and that they share β2 C.M. Moreover, relation a) shows that each pole of f is a pole of f of same order. Then by Theorem 6.4 we have f ≡ g. Suppose case b) is satisfied. Since f and g share Y2 I.M., thanks to Hypothsis (H), we can check again that f (x) = β1 if and only if g(x) = β2 and similarly, f (x) = β2 if and only if g(x) = β1 . Next, by Lemmas 2, 4, 5, f takes every value infinitely many times. Hence there exists u ∈ IK (resp. u ∈ S, resp. u ∈ D) such that f (u) = β1 and hence g(u) = β2 . But then, by relation b), we have k 

k 2   2 (β1 − αj ) = (β2 − αj )

j=1

j=1

a contradiction. Suppose finally that case c) occurs: this is exactly symmetric to case b) by permuting β1 and β2. Consequently, case a) must occur and therefore f = g. That finishes the proof of Theorem 6.5. Proof of Theorem 6.6. Let F (x) =

k 

(f (x) − αj ) and G(x) =

j=1

and γ2 =

k 

k 

(g(x) − αj ), let γ1 =

j=1

k 

(β1 − αj )

j=1

(β2 − αj ). By hypotheses of Theorem 6.6, f and g share the set {α1 , ..., αk } I.M. and the

j=1

set {β1 , β2 } C.M. hence F and G share 0 I.M. and share the set {γ1 , γ2 } C.M. Moreover, by hypothesis, we have γ1 γ2 = 0 and γ12 = γ22 . We notice that γ1 + γ2 = 0, hence when F has a zero, since G also has a zero, F (x) + G(x) cannot be equal to γ1 + γ2 . Consequently, (1)

F (x) + G(x) ≡ γ1 + γ2 .

We will show that F and G share 0 C.M. The proof is obviously trivial when F ≡ G and the case when F or G is constant is excluded by the hypotheses f ∈ M∗ (IK) and g ∈ M∗ (IK) (resp. f ∈ Mu (S) and g ∈ Mu (S), resp. f ∈ M∗ (D) and g ∈ M∗ (D)). Thus, in the following, we assume that neither F nor G are constant and that F ≡ G. Since f and g have finitely many poles in IK (resp. in S, resp. in D), and since F and G share {γ1 , γ2 } (F (x) − γ1 )(F (x) − γ2 ) has finitely many zeros and poles which C.M., it is easily seen that the function (G(x) − γ1 )(G(x) − γ2 ) only come from the poles of f and g. Consequently, there exists a function E ∈ IK(x) such that the E(x)(F (x) − γ1 )(F (x) − γ2 ) have no zero and no pole in IK (resp. in S, resp. in D). function W (x) = (G(x) − γ1 )(G(x) − γ2 ) Consequently, W belongs to IK(x) (resp. to Mb (S), resp to Mz (D)). We must now introduce three auxiliary functions. Put Q(x) = (F (x) − G(x))(F (x) + G(x) − γ1 − γ2 ) and p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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 γ2 (γ2 − γ1 )G (x) γ1 (γ1 − γ2 )F  (x) − , F (x)(F (x) − γ1 )(F (x) − γ2 ) G(x)(G(x) − γ1 )(G(x) − γ2 )   γ2 (γ2 − γ1 )F  (x) γ1 (γ1 − γ2 )G (x) φ2 (x) = Q(x) − , F (x)(F (x) − γ1 )(F (x) − γ2 ) G(x)(G(x) − γ1 )(G(x) − γ2 )   F  (x) G (x) − . φ3 (x) = Q(x) F (x)(F (x) − γ1 )(F (x) − γ2 ) G(x)(G(x) − γ1 )(G(x) − γ2 ) We will first prove that φ1 and φ2 are not identically zero. Suppose for simplicity that f and g belong to M(IK). Indeed, suppose that φ1 ≡ 0. By (1) and due to the fact that F ≡ G, we have  φ1 (x) = Q(x)

γ2 (γ2 − γ1 )G (x) γ1 (γ1 − γ2 )F  (x) ≡ . F (x)(F (x) − γ1 )(F (x) − γ2 ) G(x)(G(x) − γ1 )(G(x) − γ2 ) Let us remark that, by Lemma 2, F takes every value infinitely many times because F has finitely many poles. Let u be a zero of F − γ1 of multiplicity j. Since F and G share {γ1 , γ2 } C.M., either u is a zero of G − γ1 of same multiplicity, or it is a zero of G − γ2 of same multiplicity. Suppose first that u is a zero of G − γ1 . By developing at u, we have F (x) − γ1 = aj (x − u)j + aj+1 (x − u)j+1 ϕ(x) and G(x) − γ1 = bj (x − u)j + bj+1 (x − u)j+1 ψ(x) with ϕ, ψ ∈ M(IK), having no pole at u. γ1 (γ1 − γ2 )F  at u is j because Then by computing we can check that the residue of F (x)(F (x) − γ1 )(F (x) − γ2 γ2 (γ2 − γ1 )G jγ2 γ2 (γ2 − γ1 ) γ1 (γ1 − γ2 ) = 1 and this of is − = because F (u)(F (u) − γ2 ) G(x)(G(x) − γ1 )(G(x) − γ2 ) γ1 G(u)(G(u) − γ2 ) γ2 γ2 = −1, hence γ12 = γ22 , a contradiction to the hypothesis. Therefore, u must be a − . So, we have γ1 γ1 zero of G − γ2 of multiplicity j. So, by developing F and G at u we obtain again expressions of the form F (x) − γ1 = aj (x − u)j + aj+1 (x − u)j+1 ϕ(x) and G(x) − γ2 = bj (x − u)j + bj+1 (x − u)j+1 ψ(x) with ϕ, ψ ∈ M(IK) having no pole at u, hence we can check that the residue at u of γ1 (γ1 − γ2 )F  F (x)(F (x) − γ1 )(F (x) − γ2 ) γ2 (γ2 − γ1 )G jγ2 jγ1 is − and this of , hence we obtain again γ12 = γ22 , the same γ2 G(x)(G(x) − γ1 )(G(x) − γ2 ) γ1 contradiction. This finishes proving that φ1 ≡ 0. And similarly, we can see that φ2 ≡ 0. We will now prove that φ3 ≡ 0. Indeed, suppose that φ3 ≡ 0. We will prove that the poles of φ3 only come from the poles of F and of G. Indeed, let a be a pole of φ3 and suppose that it is not a pole of F and G. Then it is a zero F and G or a zero of (F − γ1 )(F − γ2 ) leading to a zero of (G − γ1 )(G − γ2 ). Suppose first that a is a zero of F . Then it is also a zero of G and hence that makes a pole of order 1 or zero for G F − . F (F − γ1 )(F − γ2 ) G(G − γ1 )(G − γ2 ) is −

But since a is also a zero for F − G, it is not a pole for φ3 . Similarly, let b be a zero for F − γ1 . It is then a pole of order 1 or zero for G F − . F (F − γ1 )(F − γ2 ) G(G − γ1 )(G − γ2 ) But since F and G share {γ1 , γ2 } C.M, we can check that the function (F − G)(F + G − γ1 − γ2 ) has a zero at b, hence φ3 has no pole at b. Similarly, of course, if c is a zero of F − γ2 , then φ3 has no pole at c. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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This finishes proving that each pole of φ3 must be a pole of f or g. Next, if f and g belongs to Mu (S) or to M∗ (D) we can make the same reasoning. Thus, in each hypothesis, we have N (r, φ3 ) ≤ O(log(r)) (resp. N (r, Φ3 ) ≤ O(1), resp. NR (r, φ3 ) ≤ O(log(r))). Henceforth, we will gather the two first hypotheses: f, g ∈ M∗ (IK) and f, g ∈ Mu (S) and we will consider next the hypothesis f, g ∈ M∗ (D). Let us develop φ3 . We have:  (F − γ )(F − γ ) − (G − γ )(G − γ )  F  1 2 1 2 φ3 = (F − γ1 )(F − γ2 ) F +

 (G − γ )(G − γ ) − (F − γ )(F − γ )  G F G 1 2 1 2 = (1 − EW −1 ) + (1 − E −1 W ) . (G − γ1 )(G − γ2 ) G F G

Therefore, by Lemma 7, we have M (r, φ3 ) ≤ M (r, 1 − EW −1 ) + M (r,

F G ) + M (r, 1 − E −1 W ) + M (r, ) F G

≤ 2T (r, EW −1 ) + O(1) ≤ O(log(r)) Now since N (r, φ3 ) ≤ O(log(r)), we can conclude that T (r, φ3 ) ≤ O(log(r)) (resp. since NR (r, φ3 ) ≤ O(log(r)), we can conclude that TR (r, φ3 ) ≤ O(log(r))). Now, let us develop φ1 : we have  γ (γ − γ )(F − γ )(F − γ ) − (G − γ )(G − γ )  F  1 1 2 1 2 1 2 φ1 = (F − γ1 )(F − γ2 ) F +

 γ (γ − γ )(G − γ )(G − γ ) − (F − γ )(F − γ )  G 2 2 1 1 2 1 2 (G − γ1 )(G − γ2 ) G

F G + γ1 (γ1 − γ2 )(1 − E −1 W ) . F G Therefore, as for φ3 we have T (r, φ1 ) ≤ O(log(r)) and similarly, T (r, φ2 ) ≤ O(log(r)). = γ1 (γ1 − γ2 ))(1 − EW −1 )

We will estimate Z(r, F − γ1 ). Let u be a zero of F − γ1 of multiplicity j: either it is a zero of multiplicity j of G − γ1 or it is a zero of multiplicity j of G − γ2 because F and G share the set {γ1 , γ2 } C.M. Suppose first that u is a zero of G − γ1 of multiplicity j. As it is a zero of multiplicity j of F − γ1 , it is a zero of multiplicity at least j of F + G + γ1 + γ2 . On the other hand, by developing G F − at the point u, we can check that F (F − γ1 )(F ) − γ2 ) G(G − γ1 )(G − γ2 ) G F − F (F − γ1 )(F ) − γ2 ) G(G − γ1 )(G − γ2 ) has no pole at u because u is a pole of same order for

G F and for . F (F − γ1 )(F ) − γ2 ) G(G − γ1 )(G − γ2 )

Consequently, φ3 has a zero of order at least j at u. Suppose now that u is a zero of order j of G − γ1 . Then u is a zero of order at least j for F + G − γ1 (γ1 − γ2 )F  γ2 (γ2 − γ1 )G − has no pole γ1 − γ2 . Consider now φ1 : we can check that F (F − γ1 )(F ) − γ2 ) G(G − γ1 )(G − γ2 ) at u and hence, as previously, u is a zero of order at least j for φ1 . Consequently, if f, g belongs to M∗ (IK) or to Mu (S), we have (2)

Z(r, F − γ1 ) ≤ Z(r, φ1 ) + Z(r, φ3 ) ≤ O(log(r)). p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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Similarly and symmetrically, we obtain Z(r, F − γ2 ) ≤ Z(r, φ2 ) + Z(r, φ3 ) ≤ O(log(r)).

(3)

Consequently, by (2) and (3), and by the second Main Theorem, we have T (r, F ) ≤ Z(r, F − γ1 ) + Z(r, F − γ2 ) + N(r, F ) + O(log(r)) ≤ O(log(r)) and therefore F belongs to IK(x), a contradiction. Thus the hypothesis made above: φ3 ≡ 0 is wrong. And similarly, if f, g ∈ M∗ (D), by the same reasoning, we obtain Z R (r, F − γ1 ) ≤ ZR (r, φ1 ) + ZR (r, φ3 ) ≤ O(log(r)), and Z R (r, F − γ2 ) ≤ ZR (r, φ2 ) + ZR (r, φ3 ) ≤ O(log(r)), therefore TR (r, F ) ≤ Z R (r, F − γ1 ) + Z R (r, F − γ2 ) + N R (r, F ) + O(log(r)) ≤ O(log(r)) and hence F belongs to Mz (D), a contradiction. Thus the hypothesis made above: φ3 ≡ 0 is wrong again. Consequently, now we have F  (x) G (x) ≡ . F (x)(F (x) − γ1 )(F (x) − γ2 ) G(x)(G(x) − γ1 )(G(x) − γ2 )

(4)

Consider a zero u of F of order j which by hypothesis is also a zero of G. The development at u shows that G also admits u as a zero of order j and therefore F and G share 0 C.M. But then by Theorem 6.5, we have F ≡ G. Therefore, f and g now share the set Y1 C.M., and hence, by Theorem 6.5, we can conclude f = g. Proof of Theorems 6.7. Suppose that f n (x)f m (ax + b) and gn (x)gm (ax + b) share a rational function Q ∈ IK(x) C.M. and that f and g are two distinct meromorphic functions with finitely many poles. Set F (x) = f n (x)f m (ax + b) and G(x) = gn (x)gm (ax + b). Thus F − Q and G − Q are two meromorphic functions with finitely many poles having the same zeros with respectively the same multiplicity. F −Q is a rational function h ∈ IK(x), h ≡ 1. Consequently, Therefore, by Lemma 3, G−Q G(x) = hF (x) + (1 − h)Q(x),

(1) and of course (2)

F (x) =

Q 1 G(x) + (h − 1). h h

Applying Corollary 5.2.b, by (1) and (2) we have 1 1 Q T (r, F ) ≤ Z(r, G) + Z(r, G + (h − 1) ) + N (r, F ) + o(T (r, F )) h h h (3)

= Z(r, G) + Z(r, F ) + o(T (r, F )).

Now, by Lemma 1, we have Z(r, G) ≤ 2Z(r, g) ∀r ≥ |c| and Z(r, F ) ≤ 2Z(r, f ) ∀r ≥ |c|, hence by Lemma 6 we obtain T (r, F ) ≤ 2(T (r, g) + T (r, f )) ∀r ≥ |c|. Next, by Theorems 2.2 and 3.6, we have T (r, F ) = (n + m)T (r, f ) + O(log(r)), hence (n + m)T (r, f ) ≤ 2(T (r, g) + T (r, f )) + O(log(r)). Similarly, we have (n + m)T (r, g) ≤ 2(T (r, f ) + T (r, g)) + O(log(r)). Therefore (n + m)(T (r, f ) + T (r, g)) ≤ 4(T (r, f ) + T (r, g)) + o(T (r, F )) + o(T (r, G)) + O(log(r)). But by Theorems 2.2 o(T (r, F )) is equivalent to o(T (r, F )), o(T (r, G)) is equivalent to o(T (r, g)), therefore (n + m)(T (r, f ) + T (r, g)) ≤ 4(T (r, f ) + T (r, g)) + o(T (r, f ))) + o(T (r, g)), which implies n + m ≤ 4. Consequently, F ≡ G since n + m ≥ 5. F −Q is a Now, suppose that Q is a nonzero constant and that f and g are entire functions. Then G−Q constant λ = 1 and we have (4)

G(x) = λF (x) + Q(1 − λ), p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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and of course (5)

F (x) =

Q 1 G(x) + (λ − 1). λ λ

We can apply Theorem 2.7 on M(IK) and obtain (6)

T (r, F ) ≤ Z(r, G) + Z(r, G + λ − 1) − log(r) + O(1) = Z(r, G) + Z(r, F ) − log(r) + O(1).

On the other hand, since f, g belong to A∗ (K), by Theorems 2.2, we now have T (r, F ) = T (r, f n ) + T (r, f m (ax + b)) = (n + m)T (r, f ) + O(1) and similarly, T (r, G) = (n + m)T (r, g) + O(1) . Consequently, (n + m)(T r, f ) + T (r, g)) ≤ 4(T (r, f ) + T (r, g)) − 2 log(r) + O(1), which implies n + m < 4. Consequently, if f and g belong to A∗ (IK), if Q is a nonzero constant and if n + m ≥ 4, then F ≡ G. f (x) and ψ(x) = φ(ax + b). So, we have proven that in each case, we have F ≡ G. Now, set φ(x) = g(x) 1 Then we have φn (x)ψ m (x) ≡ 1, hence m ≡ φn and hence ψ (7)

T (r,

1 ) = T (r, φn ). ψm

But T (r, ψ1 ) = T (r, ψ) = T (r, φ) ∀r ≥ |c|. On the other hand, by Theorems 2.2 T (r, φn ) = nT (r, φ), T (r, ψ m ) = mT (r, ψ) hence by (7), since m = n, we can see that T (r, φ) = 0. Therefore, φ is a constant t and hence f = tg. Now, F − Q belongs to M∗ (IK), has finitely many poles and, by Lemma 2, admits infinitely many zeros that are not zeros of Q. Let c be such a zero. Then c is also a zero of G − Q and hence we have f n (c)f m (ac + b) = gn (c)gm (ac + b), therefore f n (c)f m (ac + b) = tn f n (c)tm f m (ac + b) i.e. F (c) = tn+m F (c). But since c is not a zero of Q, we have F (c) = 0, hence tn+m = 1, which ends the proof. Proof of Theorems 6.8. The proof of Theorem 6.8 roughly follows that of Theorem 6.7. Suppose first that f n (x)f m (ax + b) and gn (x)gm (ax + b) share C.M. a function θ ∈ M(S) (resp. θ ∈ M(D)) having finitely many zeros and poles in S (resp. in D), and are two distinct functions lying in Mu (S) (resp. in M∗ (D)) and have finitely many poles in S (resp. in D). Set again F (x) = f n (x)f m (ax + b) and G(x) = g n (x)gm (ax + b). By Lemma 4, F − θ and G − θ belong to Mu (S) (resp. to M∗ (D)) and are two functions having the same zeros with respectively the same multiplicity and have finitely many poles. F −θ belongs to M(S) (resp. to M(D)) and has finitely many zeros and poles in S (resp. in Therefore G−θ D). Consequently, it is a function w that belongs to Mb (S) (resp. to Mz (D)). Suppose first f, g ∈ Mu (S). Then by Corollary 5.2.b, similarly to the proof of Theorem 6.7, we obtain T (r, F ) ≤ Z(r, G) + Z(r, G + ( (1)

1 − 1)θ + o(T (r, F )) (r → R) = Z(r, G) w

+Z(r, F ) + o(T (r, F )) (r → R).

Now, by Lemma 1, we have again Z(r, G) ≤ 2Z(r, g) + O(1) and Z(r, F ) ≤ 2Z(r, f ) + O(1), hence by Lemma 6 we obtain T (r, F ) ≤ 2(T (r, g) + T (r, f ) + O(1) hence, by Theorems 2.2, (n + m)T (r, g) ≤ 2(T (r, g) + T (r, f )) + O(1) and similarly (n + m)T (r, f ) ≤ 2(T (r, f ) + T (r, g)) + O(1). On the other hand, o(T (r, F )) = o(T (r, f )), o(T (r, G)) = o(T (r, g)), therefore by (1), we obtain (n + m)(T r, f ) + T (r, g)) ≤ 4(T r, f ) + T (r, g)) + o(T (r, f )) + o(T (r, g)) and hence n + m < 5, which is excluded by hypotheses. Consequently, we get F ≡ G. Putting again 1 1 f (x) and ψ(x) = φ(ax + b), we have m ≡ φn , therefore in S we get T (r, m ) = T (r, φn ) φ(x) = g(x) ψ ψ which implies that T (r, φ) = 0 since n = m. Thus φ is a constant t again and hence, since θ has finitely many zeros in S, by Lemma 4 we can conclude as in Theorem 6.7. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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Similarly, if f, g ∈ M∗ (D) we can make the same reasonning thanks to Lemma 5. Proof of Theorem 6.9. Suppose first that f belongs to M∗ (IK) and w ∈ M(IK) and set h(x) = f (ax + b), F (x) = f (x)n hm (x). By Corollary 5.2.a we have T (r, F ) ≤ Z(r, F ) + Z(r, F − w) + N (r, F ) + o(T (r, F )) (r → +∞).

(1) Clearly,

Z(r, F ) ≤ Z(r, f ) + Z(r, h),

(2) and (3)

N (r, F ) ≤ N (r, f ) + N (r, h) + N (r, w) = N (r, f ) + N (r, h) + o(T (r, f )) (r → +∞).

On the other hand, by Theorems 2.2, T (r, f n hm ) ≥ |T (r, f n ) − T (r, hm )|∞ and by Lemma 6, for every r ≥ |b| we have T (r, h) = T (r, f ), Z(r, f ) = Z(r, h), N (r, f ) = N (r, h), hence T (r, F ) ≥ |n − m|∞ T (r, f ). Moreover, o(T (r, F )) = o(T (r, f )). Therefore, by (1), (2) and (3) we obtain (|n − m|∞ − 4)T (r, f ) ≤ Z(r, F − w) + o(T (r, f )) (r → +∞) which proves the claim whenever |n − m|∞ ≥ 5. Suppose now that f belongs to Mu (S) and w ∈ M(S). Keeping the same notations, we can get the same relations (1), (2), (3) and next (|n − m|∞ − 4)T (r, f ) ≤ Z(r, F − w) + o(T (r, f )) (r → R), which proves again the claim whenver |n − m|∞ ≥ 5. Finally, suppose that f belongs to M∗ (D) and w ∈ M(D) and keep the same notations. By Corollary 5.2.a now we have TR (r, F ) ≤ Z R (r, F ) + Z R (r, F − w) + N R (r, F ) + o(T (r, F )).

(4)

Next, using Theorem 3.6 instead of 2.2, we obtain here: (|n − m|∞ − 4)TR (r, f ) ≤ Z R (r, F − w) + o(TR (r, f )) (r → +∞). Thus, if |n − m|∞ ≥ 5 we have TR (r, f ) ≤ Z R (r, F − w) + o(TR (r, f )) (r → +∞) which shows again that F − w has infinitely many zeros . Proof of Theorem 6.10. The beginning of the proof is similar to that of Theorem 6.9. Suppose first that f ∈ M∗ (IK) and set h(x) = f (ax + b), F (x) = f (x)n hm (x). Suppose that F admits 4 distinct perfectly branched values a1 , a2 , a3 , a4 . By Theorem 2.7, we have 3T (r, F ) ≤

4 

Z(r, F − aj ) + N (r, F ) − log(r) + O(1) (r → +∞),

j=1

hence (1)

4 1  Z(r, F − aj ) + N (r, f ) + N (r, h) − log(r) + O(1) (r → +∞), 3T (r, F ) ≤ 2 j=1

and hence 3(|n − m|∞ )T (r, f ) ≤

1 4(n + m)T (r, f )) + 2T (r, f ) + O(log(r)) (r → +∞), 2

therefore 3|n − m|∞ ≤ 2(n + m) + 2 + λ(r), with lim λ(r) = 0. Consequently, if 3|n − m|∞ > 2(n + r→+∞

m + 1), the function F does not admit 4 perfectly branched values. Moreover, if F admits 4 totally branched values a1 , a2 , a3 , a4 , then by (1) now we can get 1 4(n + m)T (r, f )) + 2T (r, f ) − log(r)) + O(1) (r → +∞), 3(|n − m|∞ )T (r, f ) ≤ 2 therefore 3|n − m|∞ < 2(n + m) + 2. Consequently, if 3|n − m|∞ ≥ 2(n + m + 1), the function F does not admit 4 totally branched values. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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Suppose now that F admits 3 distinct perfectly branched values a1 , a2 , a3 . By Theorem 2.7, we have 2T (r, F ) ≤

(2)

3 

Z(r, F − aj ) + N (r, F ) − log(r) + O(1) (r → +∞),

j=1

1 3(n + m)T (r, f )) + 2T (r, f ) + λ(r) 2   lim λ(r) = 0, (r → +∞), therefore 2|n − m|∞ ≤ 32 (n + m) + 2. Consequently, if 4|n −

and hence 2(|n − m|∞ )T (r, f ) ≤ with

r→+∞

m|∞ > 3(n + m) + 4, the function F does not admit 3 perfectly branched values. Moreover if F admits 3 totally branched values a1 , a2 , a3 , then by (2) now we can get 1 3(n + m)T (r, f )) + 2T (r, f ) − log(r) + O(1) (r → +∞), 2(|n − m|∞ )T (r, f ) ≤ 2 therefore 4|n − m|∞ < 3(n + m) + 4. Consequently, if 4|n − m|∞ ≥ 3(n + m) + 4, the function F does not admit 3 totally branched values. Finally, if f belongs to Mu (S) or to M∗ (D), using in each case Theorem 2.7 or 3.10, we can do the same reasoning for perfectly branched values. But Theorem 5.2 here does not show the term − log(r) + O(1) which is essential when we consider totally branched values. This is why we must replace ≥ by >in each statement for functions in Mu (S) or in M∗ (D) concerning totally branched values. ACKNOWLEDGEMENTS We are grateful to the referee for pointing out to us many misprints. The second named author is supported by Vietnam’s National Foundation for Science and Technology Development (NAFOSTED) under grant number 101.04-2017.320 and Vietnam Institute Advanced Study in Mathematics. REFERENCES 1. T. T. H. An, “Defect relation for non-Archimedean analytic maps into arbitrary projective varieties,” Proc. Amer. Math. Soc. 135, 1255–1261 (2007). 2. K. Boussaf, “Motzkin factorization in algebras of analytic elements,” Ann. Math. Blaise Pascal 2 (1), 73–91 (1995). 3. K. Boussaf, A. Escassut and J. Ojeda, “Complex and p-adic branched functions and growth of entire functions,” Bull. Belgian Math. Soc. Simon Stevin 22, 781–796 (2015). ´ 4. A. Boutabaa, “Theorie de Nevanlinna p-adique,” Manuscripta Math. 67, 251–269 (1990). 5. A. Boutabaa and A. Escassut, “URS and URSIMS for p-adic meromorphic functions inside a disk,” Proc. Edinburgh Math. Society 44, 485–504 (2001). 6. K. S. Charak, “Value distribution theory of meromorphic functions,” Math. Newsletter 18 (4), 1–35 (2009). 7. J. F. Chen, “Uniqueness of meromorphic functions sharing two finite sets,” paper to appear. 8. A. Escassut, Value Distribution in p-Adic Analysis (WSCP Singapore, 2015). 9. A. Escassut and T. T. H. An, “p-Adic Nevanlinna theory outside of a hole,” Vietnam J. Math. 45 (4), 681–694 (2017). 10. H. K. Ha, “On p-adic meromorphic functions,” Duke Math. J. 50, 695–711 (1983). 11. M. O. Hanyak and A. A. Kondratyuk, “Meromorphic functions in m-punctured complex planes,” Matematychni Studii 27 (1), 53–69 (2007). 12. Vu Hoai An, Pham Ngoc Hoa and Ha Huy Khoai, “Value sharing problems for differential polynomials of meromorphic functions in a non-Archimedean field,” paper to appear. ´ complets,”Les ten13. M. Krasner, “Prolongement analytique uniforme et multiforme dans les corps values dances geom ´ etriques ´ en algebre ` et theorie ´ des nombres, Clermont-Ferrand, pp.94–141 (1964). Centre National de la Recherche Scientifique (1966). (Colloques internationaux de C.N.R.S. Paris, 143). 14. Liu Gang and Meng Chao, “Uniqueness for the difference monomials of p-adic entire punctions,” paper to appear. ´ ´ 15. E. Motzkin, “La decomposition d’un el´ ement analytique en facteurs singuliers,” Ann. Inst. Fourier 27 (1), 67–82 (1977). 16. X. G. Qi, L. Z. Yang and K. Liu, “Uniqueness and periodicity of meromorphic functions concerning the difference operator,” Comput. Math. Appl. 60, 1739–1746 (2010). 17. K. Yamanoi, “The second main theorem for small functions and related problems,” Acta Math. 192, 225–294 (2004). p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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